Question

x la 2 plus 5x minus 6

x la 2 minus x minus 6
X la 2 plus 11x minus 12
X la 2-6x-27
(x la 2+8)(x la 2+x-8)+12
X(x+4)(x la 2+4x+6)+8
VA ROG SA ÎMI REZOLVĂRI ACESTE EUCUATI, DAU COROANA, MULȚUMIRI ANTICIPATE ​

Answer 1

Presupun ca toate ecuatiile sunt egale cu 0.

1) x²+5x-6=0

   Calculam Δ=b²-4ac => Δ=25+24=49 => √Δ=T=7

x1=$\frac{-b+T}{2}$=$\frac{-5+7}{2}$=1

x2=$\frac{-b-T}{2}$=$\frac{-5-7}{2}$=-6

2) x²-x-6=0

   Calculam Δ=b²-4ac => Δ=1+24=25 => √Δ=T=5

x1=$\frac{-b+T}{2}$=$\frac{1+5}{2}$=3

x2=$\frac{-b+T}{2}$=$\frac{1-5}{2}$=-2

3) x²+11x-12=0

Ecuatia mai poate fi scrisa si asa: x²+12x-x-12=0, grupand cate 2 termenii, ne rezulta: x(x+12)-(x+12)=0=> (x-1)(x+12)=0 => x1=1, x2=-12.

4) x²-6x-27=0

Calculam Δ=b²-4ac => Δ=36+108=144 => √Δ=T=12

x1=$\frac{-b+T}{2}$=$\frac{6+12}{2}$=9

x2=$\frac{-b-T}{2}$=$\frac{6-12}{2}$=-3

5) (x²+8)(x²+x-8)+12=0

Cred ca ai gresit aici, ecuatia nu are solutii reale sau complexe.

6) x(x+4)(x²+4x+6)+8=0

=> $x^{4}+8x^{3}+22x^{2} +24x+8=0$

Scriem ecuatia sub aceasta forma:

$x^{4} +2x^{3}+6x^{3}+12x^{2}+10x^{2}+20x+4x+8=0$

=> $x^{3}(x+2)+6x^{2}(x+2)+10x(x+2)+4(x+2)=0$

=> $(x+2)(x^{3}+6x^{2}+10x+4)=0$

=>$(x+2)(x^{3}+2x^{2}+4x^{2}+8x+2x+4)=0$

=> $(x+2)(x^{2}(x+2)+4x(x+2)+2(x+2))$

=> $(x+2)^{2} (x^{2}+4x+2)=0$

De aici ne rezulta ca x1=x2=-2 si ca x²+4x+2=0.

Rezolvam ultima ecuatie:

Calculam Δ=b²-4ac => Δ=16-8=8 => √Δ=T=2√2

x3=$\frac{-b+T}{2}$=$\frac{-4+T}{2}$=-2+√2

x4=$\frac{-b-T}{2}$=$\frac{-4-T}{2}$=-2-√2